Exercise 10, § 8
Tag 02HO, permanent. Cite it as [Bourbaki, Tag 02HO].
Spectral Theories, I, p. 195
Bourbaki marks this one as harder than the rest.
Let X be a compact topological space and A a unital Banach subalgebra of for the induced norm.
a) A subset K of X is said to be antisymmetric (relative to A) if every function that is real on K is constant on K. Every antisymmetric subset is contained in a maximal antisymmetric subset. A maximal antisymmetric subset is closed. The set of maximal antisymmetric subsets is a partition of X.
b) Let be the orthogonal of A in the Banach space of complex measures on X. Let be an extreme point of the unit ball of . Now prove that the support of is antisymmetric.
c) Let . If for every , then . (Apply b) and the Krein–Milman theorem.) Hence recover the Weierstrass–Stone theorem (TG, X, p. 36, th. 3).
d) A nonempty subset E of X is called a peak (relative to A) if there exists such that and E is the set of for which . One may then suppose that for (replace by ). Every nonempty countable intersection of peaks is a peak. If are peaks, then is a peak. (Let with on on X - ; let ; then on on X - .)
e) Let E be an intersection of peaks. Let I be the set of vanishing on E. Then defines an isometry of onto the algebra of restrictions to E of functions in A.
f) Let E be a peak. Let be a peak relative to . Then is a peak relative to A.
g) Every antisymmetric subset of X is an intersection of peaks. (Use d) and f).) h) For every antisymmetric subset K of X, the algebra is a closed subalgebra of . (Use e) and g).)
i) Let R be the set of real parts of the elements of A. Suppose that and that R is stable under multiplication. Then X is reduced to a point. (Let . For , there exists a unique such that and ; set . Then R is a real Banach space for N; the closed graph theorem proves that multiplication in R is separately continuous, hence continuous. Deduce on a Banach algebra structure. Let with for every . Now prove that, for every character of S, one has , and deduce that log(). Assuming , one can construct an such that on X and such that is arbitrarily large. From the above, there exists with . One has therefore , but
1
2
is arbitrarily large.)
j) If R is stable under multiplication, then . (Use a), c), h), i).) Consequently, if , there exists such that .
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